Problem: Nauuo is a girl who loves playing chess.

One day she invented a game by herself which needs n chess pieces to play on a m× m chessboard. The rows and columns are numbered from 1 to m. We denote a cell on the intersection of the r-th row and c-th column as (r,c).

The game's goal is to place n chess pieces numbered from 1 to n on the chessboard, the i-th piece lies on (r_i,\,c_i), while the following rule is satisfied: for all pairs of pieces i and j, |r_i-r_j|+|c_i-c_j|≥|i-j|. Here |x| means the absolute value of x.

However, Nauuo discovered that sometimes she couldn't find a solution because the chessboard was too small.

She wants to find the smallest chessboard on which she can put n pieces according to the rules.

She also wonders how to place the pieces on such a chessboard. Can you help her?

Input

The only line contains a single integer n (1≤ n≤ 1000) — the number of chess pieces for the game.

Output

The first line contains a single integer — the minimum value of m, where m is the length of sides of the suitable chessboard.

The i-th of the next n lines contains two integers r_i and c_i (1≤ r_i,c_i≤ m) — the coordinates of the i-th chess piece.

If there are multiple answers, print any.

Examples

Input


2


Output


2
1 1
1 2

Input


4


Output


3
1 1
1 3
3 1
3 3

Note

In the first example, you can't place the two pieces on a 1×1 chessboard without breaking the rule. But you can place two pieces on a 2×2 chessboard like this:

<image>

In the second example, you can't place four pieces on a 2×2 chessboard without breaking the rule. For example, if you place the pieces like this:

<image>

then |r_1-r_3|+|c_1-c_3|=|1-2|+|1-1|=1, |1-3|=2, 1<2; and |r_1-r_4|+|c_1-c_4|=|1-2|+|1-2|=2, |1-4|=3, 2<3. It doesn't satisfy the rule.

However, on a 3×3 chessboard, you can place four pieces like this:

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const long double pie = 3.1415926535;
using namespace std;
long long fpow(long long n, long long p) {
  ;
  long long m = 1;
  while (p) {
    if (p % 2) m *= n;
    p >>= 1;
    n *= n;
  }
  return m;
}
long long mfpow(long long n, long long p, long long M) {
  long long m = 1;
  n %= M;
  while (p) {
    if (p % 2) m = (m * n) % M;
    n = (n * n) % M;
    p >>= 1;
  }
  return m % M;
}
long long min() { return 1e18; }
long long min(long long a) { return a; }
long long min(long long a, long long b) {
  if (a < b)
    return a;
  else
    return b;
}
template <typename... Args>
long long min(long long a, long long b, Args... args) {
  if (a < b)
    return min(a, args...);
  else
    return min(b, args...);
}
long long max() { return -1e18; }
long long max(long long a) { return a; }
long long max(long long a, long long b) {
  if (a < b)
    return b;
  else
    return a;
}
template <typename... Args>
long long max(long long a, long long b, Args... args) {
  if (a < b)
    return max(b, args...);
  else
    return max(a, args...);
}
const vector<long long> days = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
bool br1, br2, br;
char ch, ch1, ch2;
long long n, k, x, y, z, mini, maxi, l, ind, ini, sum, t, len, r, w, imini, m;
pair<long long, long long> p;
string s, s1, s2;
void solve() {
  cin >> n;
  for (long long i = 1; i < 1e5; i++) {
    if (2 * i - 1 >= n) {
      cout << i << endl;
      x = 1;
      y = 1;
      for (long long j = 0; j < n; j++) {
        cout << x << " " << y << endl;
        if (y == i)
          x++;
        else
          y++;
      }
      break;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
  return 0;
}